The standard equation for an hyperbola with axes paralleling the x- and y-axes and center at the point (h, k) is as follows:
. . . . .\(\displaystyle \large{\frac{(x\,-\,h)^2}{a^2}\,-\,\frac{(y\,-\,k)^2}{b^2}\,=\,1}\)
...if the major axis (the one containing the vertices and the center) is parallel to the x-axis, and:
. . . . .\(\displaystyle \large{\frac{(y\,-\,k)^2}{a^2}\,-\,\frac{(x\,-\,h)^2}{b^2}\,=\,1}\)
...if the major axis is parallel to the y-axis. That is, the "a" value (with the major axis having length 2a) always goes with the variable whose axis the major axis parallels. The "b" value (for the other, "conjugate", axis of length 2b) goes with the other variable. The variable being subtracted is always the minor-axis variable.
Since the conjugate axis in this case has length 12 = 2b, then b = 6. Since the vertices lie on the major axis and form its endpoints, then the major axis has length 10 = 2a, and a = 5. Also, the center is the midpoint of the major axis, so (h, k) = (0, 0). (This is why the equation simplified to that which was posted earlier: since h = k = 0, the numerators of the fractions simplified to just being the squares of the variables x and y.)
Since the vertices are placed horizontally rather than vertically, then the major axis parallels the x-axis, not the y-axis. This should be enough information to allow solution of this and any similar exercises.
Eliz.
